Given a sequence of real numbers,do the following conditions suffice to guarantee convergence to 0?

Question

If $x_{a+1}$-$x_{a}$ converges to $0$ and $x_{2a}$-$2x_{a}$ converges to $0$ , does that imply $x_a$ converges to $0$?

Answer 1

Yes. There's probably a clever proof, but here's a non-clever one. Suppose you had a sequence satisfying your hypotheses but not converging to 0. Multiplying it by a suitable constant (which doesn't affect the hypotheses), you can assume that $x_a>1$ for infinitely many $a$, in particular for some $a$ so large that $|x_{2b}-2x_b|<\frac12$ for all $b\geq a$. Applying that inequality repeatedly, with $a,2a,4a,\dots$ as $b$, you get that $x_{2^ka}>(2^k+1)/2$ for all integers $k\geq0$. Then \[ x_{2^{k+1}a}-x_{2^ka}\geq x_{2^ka}-\frac12\geq2^{k-1}. \] That is, as $b$ increases from $2^ka$ to $2^{k+1}a$, a distance of $2^ka$, the value of $x_b$ increases by at least $2^{k-1}$. Therefore, at some step along the way, from some $b$ to $b+1$, the value of $x_b$ must have increased by at least $1/(2a)$. This happens repeatedly, with the same $a$, because $k$ can increase arbitrarily, and it therefore contradicts the hypothesis that $x_{b+1}-x_b$ tends to 0.

Answer 2

Just a very quick argument which reduces the possibilities: Let $\Omega\subset{\mathbb R}\cup\{\pm\infty\}$ be the $\omega$-limit set of the sequence, that is the set of limits of "converging" sub-sequences. It is a non-void closed set by construction. The property $x_{a+1}-x_a\rightarrow0$ tells us that $\Omega$ is a connected set. The property $x_{2a}-2x_a\rightarrow0$ tells us that $2\Omega=\Omega$. Therefore $\Omega$ can only be equal to one of the four sets $$\{0\},\quad[0,+\infty],\quad[-\infty,0],\quad{\mathbb R}.$$

Edit. It was commented that the second property gives only an inclusion, of $2\Omega$ into $\Omega$. Actually, it does give also the reverse inclusion (hence the equality), when combined with the first property: Let $\ell$ be the limit of some subsequence $x_{n_k}$. Because of the first property, we may suppose that $n_k=2m_k$ is even. Then $\ell/2$ is the limit of $x_{m_k}$, hence $\ell/2\in\Omega$.